Tangent $\overline{TB}$ and secant $\overline{TCA}$ are drawn to circle $O$. Diameter $\overline{AB}$ is drawn. If $TC = 6$ and $CA = 10,$ then $CB =$

Because AB is the diameter, the angle in the semicircle is a right angle, so triangle ABC is a right-angled triangle with the right angle at C. Using the Pythagorean theorem, CB^2 + CA^2 = AB^2. Substituting the given value CA = 10 and finding AB as the sum of the secant segments TC + CA = 6 + 10 = 16, we get CB^2 + 10^2 = 16^2. Therefore, CB^2 = 256 - 100 = 156, so CB is the square root of 156, which simplifies to 2 times the square root of 15.

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